From Wikipedia:

a uniform space is called complete if every Cauchy filter converges.

  1. I was wondering if the following three are equivalent in a uniform space:

    Or, which one implies which but doesn't imply which? For example, are the first two equivalent, while the third is implied by but does not implie any of the first two?

  2. How about in a metric space?

Thanks and regards!


In a metric space all three properties are equivalent.

In a uniform space every Cauchy filter converges iff every Cauchy net converges; the usual equivalence between filters and nets in arbitrary topological spaces preserves the property of being Cauchy in uniform spaces. This is strictly stronger than merely requiring Cauchy sequences to converge.

Example: Let $X$ be $\omega_1$ with the order topology. $X$ is Tikhonov, so it has a compatible uniformity $\mathscr{U}$. $X$ is countably compact, so $\langle X,\mathscr{U}\rangle$ is totally bounded, but $X$ is not compact, so $\langle X,\mathscr{U}\rangle$ is not complete: a uniform space is compact iff it is complete and totally bounded. Thus, $\langle X,\mathscr{U}\rangle$ must have Cauchy filters/nets that do not converge. However, every sequence in $X$ is contained in a compact subspace of $X$, so every Cauchy sequence does converge.

  • 1
    $\begingroup$ "In a metric space all three properties are equivalent" is equivalent to Countable Choice $\hspace{1 in}$ (en.wikipedia.org/wiki/Axiom_of_countable_choice). $\;$ $\endgroup$ – user57159 Feb 2 '12 at 3:57
  • $\begingroup$ +1, thanks! In a topological vector space: (1) are the three equivalent, and (2) are convergent filter, convergent net and convergent sequence equivalent? $\endgroup$ – Tim Feb 2 '12 at 8:17
  • $\begingroup$ @RickyDemer: How is it equivalent to countable choice? $\endgroup$ – Tim Feb 2 '12 at 8:35
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    $\begingroup$ @Tim: I don’t offhand see how to do it for metric spaces, but for pseudometric spaces it’s not too hard. If CC fails, one can construct a pseudometric space $X=\bigsqcup_nX_n$ that has no Cauchy sequences (so vacuously all Cauchy sequences converge!) but is such that $\mathscr{F}=\{\bigcup_{k\ge n}X_k:n\in\omega\}$ is a Cauchy filter; clearly $\mathscr{F}$ does not converge. $\endgroup$ – Brian M. Scott Feb 2 '12 at 22:06
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    $\begingroup$ @Tim, a topological vector space is a uniform space in particular, so nets and filters are equivalent (for Cauchyness) and still sequences do not suffice in general $\endgroup$ – Henno Brandsma Feb 5 '12 at 20:36

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